Sir I am from India Do you know about JEE Advanced exam of India

if f(f(y))=y and f is surjective does that also mean f(y)=y?

Where are you from?

Oh dear Lord, your channel is EXACTLY what I needed to find. Thank you so much for this, I'll be bingingnthis series and taking extensive notes prepping for math olympiads. Hope to see y'all at the IMO!

In the step where you conclude that 1=f(f(y))+f(y). Would it be possible to make a substitution such that u=f(y)? Which would make the equation 1=f(u)+u. This can then be rearranged to become f(u)=1-u.

How can we know that this is the only solution?

Hello sir, please do you offer 1 to 1 online lessons ?

Is there another word for the circle method? I have tried to search for it on Google but I didn’t get any results

THANK YOU!!! All the other videos on here only talked about how to prove a function is surjective, and not how to use it’s properties to solve problems. This really helped me

Thank you!

Very fun and easy problem!

thank you!

Great🎉

hi sir please can you solve this functional equation i try for more times but i can't solve it f(xf(y)-f(x))=2f(x)+xy

Hey sir, I was wondering what I should do in order to prepare for math Olympiad. Im currently year 10 (gcse) and can u please recommend a few textbooks for the preparation.

Solution 3: Let H be the orthocentre of ABC. Then HBX=HBA=HCA=HCX, so HBCX concyclic. This gives BXC=BHC=180-BAC=180-BZC=BYC. Also XCB=YCB and XB=YB, so triangles XCB and YCB are congruent. That gives BC perpendicular to XY, so BC perpendicular to ZX. Combining this with BXC=180-BZC implies that X is orthocenter of BCZ, so BZ perpendicular to CX

Another solution: B is on perp bisector of AY, so on angle bisector of AZY. So B is on angle bisector of AZX and on perp bisector of AX. But obviously ABXZ aren’t concyclic since X!=C. This means that the angle bisector of AZX is the same line as the perp bisector of AX. So Z is on perp bisector of AX. Since B is also onnthat perp bisector we have BZ perpendicular to AC

Solved in 1 minute, solution: angle chase BZX+ZXA=BZY+180-AXY=BAY+ABY/2=90, so BZ is perpendicular to AC

Well I came up with a difference solution using both "bad points". So let D be the intersection of |AY| and |BZ|. Notice that since |BY|=|BA| we have that B is the mid point of the arc AY so BZ is the angle bisector of [angle AZY] and used this with the fact that ABCZY is cyclic we'll have [angle AZB]=[angle YZB]=[angle BYA]=[angle BAY]. Now notice that ∆BYD~∆BZY because [angle BYD]=[angle BZY] and they have a common angle at B therefore [angle BDY]=[angle BYZ] but the ∆BXY is isosceles so [angle BYZ]=[angle BYX]=[angle BXY]=[angle BDY] so the quadrilateral BYXD is cylic thefore [angle DYX]=[angle XBD]. With this we have [angle AYZ]=[angle ABZ]=[angle XBD] so BZ is the angle bisector of [angle XBA ] and since ∆BXA is isosceles it is also it's perpendicular bisector so that finishes the problem.

Thanks for the video. Here a solution involving the bad points X and Y: As |AB| = |BY|, one has that [angle AZB] = [angle ACB] = [angle BZY] = [angle BCY]. In the circle with center at B and passing through A, X and Y, one has that [angle AXY] + [angle ABY]/2 = 180°, which means that [angle AXZ] = [angle ABY]/2. As the points B and Z lie on opposite sides of AY, one has that [angle ABY] + [angle AZY] = 180° = 2*[angle AXZ] + 2*[angle BZY], i.e. [angle AXZ] + [angle BZY] = 90°, which means that AC and BZ are perpendicular.

Lets go!!!

I am going to this year's IMO any advices for the contest?

this was fairly easy problem

Its a good day when little fermat uploads

I found the answer graphically in less than 1 minutes😭 1) the function is odd and therefore its graph is symmetrical about the origin 2) for all values of x, when we increase X by 1 f(x) is also increased by 1, therefore it had to be a line of gradient/slope 1, only the graph of y=X could hold 3) just checked if f(x)=X was true for the last one, And bam!, I know it just worked for that particular case but at that instant of my life, I felt so smart😭😭 It gives me hope Thanks so much for your videos ❤️!

We had this as P2 last year during nationals with the word KLAUSUR(German for exam)

discontinuing inequalities?

Great video. In fact, the identities are valid for x different from 0 and 1. You can check that for all real numbers a, the piecewise function defined by f(x) = {x if x<>0 / a if x=0 / a+1 if x=1 satisfies the functional equation.

Thanks for the awesome video! I always love your content. Do you happen to offer any online coaching for the IMO? Love from Syria 🤍

How do you write pie -2x , plz explain

Next inequality? Cauchy?

Thnx❤

we deserve a projective geometry course bro. give that as an early Christmas gift.

Thanks bro

thank you so much <3

Hi can you contact with me an important matter?

Is functional equations your favourite?

I was looking for this, thank you❤.

Op 🔥🔥

Hello sir can you tell me where are you from?